What is the integral of (x^2+15)(x^4-9) ?

The integral of (x^2+15)(x^4-9) is (x^7)/7-3x^3+3x^5-135x+C

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integral of (x^2+15)(x^4-9)

\int (x^{2} + 15) (x^{4} - 9) d x

\frac{x ^{7}}{7} - 3 x^{3} + 3 x^{5} - 135 x + C

Solution steps

\int (x^{2} + 15) (x^{4} - 9) d x

Expand

(x^{2} + 15) (x^{4} - 9) : x^{6} - 9 x^{2} + 15 x^{4} - 135

= \int x^{6} - 9 x^{2} + 15 x^{4} - 135 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{6} d x - \int 9 x^{2} d x + \int 15 x^{4} d x - \int 135 d x

\int x^{6} d x = \frac{x ^{7}}{7}

\int 9 x^{2} d x = 3 x^{3}

\int 15 x^{4} d x = 3 x^{5}

\int 135 d x = 135 x

= \frac{x ^{7}}{7} - 3 x^{3} + 3 x^{5} - 135 x

Add a constant to the solution

= \frac{x ^{7}}{7} - 3 x^{3} + 3 x^{5} - 135 x + C

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What is the integral of (x^2+15)(x^4-9) ?
The integral of (x^2+15)(x^4-9) is (x^7)/7-3x^3+3x^5-135x+C